Appendix - Mathematical Induction and Other Forms of Proofs - Exercises - Page A6: 22

See the proof below.

Let $S=\left\{\textbf{u},\textbf{v}\right\}$ be a linearly independent set in some vector space $V$ and let $S'=\left\{\textbf{u}+\textbf{v},\textbf{u}-\textbf{v}\right\}$. We will prove by contradiction that $S'$ is linearly independent. $\textbf{Proof:}$ Suppose that $S'$ is a linearly dependent set. Then, there exists scalars $a$ and $b$ not both $0$ such that $$a\left(\textbf{u}+\textbf{v}\right)+b\left(\textbf{u}-\textbf{v}\right)=\textbf{0}.$$ Now, since $$a\left(\textbf{u}+\textbf{v}\right)+b\left(\textbf{u}-\textbf{v}\right)=a\textbf{u}+a\textbf{v}+b\textbf{u}+b\left(-\textbf{v}\right)=$$ $$a\textbf{u}+b\textbf{u}+a\textbf{v}+b\left(\left(-1\right)\textbf{v}\right)=\left(a+b\right)\textbf{u}+a\textbf{v}+\left(b\left(-1\right)\right)\textbf{v}=$$ $$\left(a+b\right)\textbf{u}+a\textbf{v}+\left(-b\right)\textbf{v}=\left(a+b\right)\textbf{u}+\left(a+\left(-b\right)\right)\textbf{v}=$$ $$\left(a+b\right)\textbf{u}+\left(a-b\right)\textbf{v}$$ we have that $$\left(a+b\right)\textbf{u}+\left(a-b\right)\textbf{v}=\textbf{0}.$$ _____________________________________________________________________________ Let's take a look at the linear system $$\left\{\begin{array}{cc} c_1+c_2=0 \\ c_1-c_2=0\end{array}\right..$$ Adding the two equations in the system we obtain that $c_1+c_2+c_1-c_2=0$ which gives us that $2c_1=0$ and therefore, $c_1=0$. From $c_1+c_2=0$ and $c_1=0$ we obtain that $0+c_2=0$ and therefore, $c_2=0$. Thus, we have shown that the linear system $\left(0,0\right)$ is the only solution of the system above. Thus, since $a$ and $b$ are not both $0$, we have that $\left(a,b\right)$ is not a solution of the system which implies that $a+b$ and $a-b$ cannot both be $0$ and then, since $$\left(a+b\right)\textbf{u}+\left(a-b\right)\textbf{v}=\textbf{0}$$ this contradicts the fact that $S$ is linearly independent. Hence, $S'$ is a linearly independent set.